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Use the Randomization test for two proportions applet to approximate the P-value and decide whether to reject or not reject [tex]H_0[/tex]. Use the count of [tex]x_1[/tex] as the test statistic. Conduct 2000 random assignments. Use the seed value 12641 and a significance level of [tex]\alpha=0.05[/tex].

[tex]H_0: p_1=p_2[/tex] versus [tex]H_1: p_1\ \textgreater \ p_2[/tex]

Sample data: [tex]x_1=359, n_1=546, x_2=359, n_2=584[/tex]

To open the Randomization test for two proportions applet in StatCrunch, select Applets, highlight Resampling, then select Randomization test for two proportions.

The estimated P-value using 2000 random assignments is [tex]\square[/tex].

(Round to three decimal places as needed.)

Sagot :

GinnyAnsweridn
To address this problem, we need to follow the steps of conducting a randomization test for two proportions. We have the sample data: [tex]\( x_1 = 359, n_1 = 546, x_2 = 359, n_2 = 584 \)[/tex].

### Steps to Conduct the Randomization Test:

1. Formulate Hypotheses:
- Null hypothesis, [tex]\( H_0 \)[/tex]: [tex]\( p_1 = p_2 \)[/tex]
- Alternative hypothesis, [tex]\( H_1 \)[/tex]: [tex]\( p_1 > p_2 \)[/tex]

2. Set Parameters:
- Sample data: [tex]\( x_1 = 359, n_1 = 546, x_2 = 359, n_2 = 584 \)[/tex]
- Significance level: [tex]\( \alpha = 0.05 \)[/tex]
- Seed value: 12641
- Number of random assignments (iterations): 2000

3. Calculate Observed Proportion:
- Proportion [tex]\( \hat{p}_1 = \frac{359}{546} \approx 0.657 \)[/tex]
- Proportion [tex]\( \hat{p}_2 = \frac{359}{584} \approx 0.615 \)[/tex]

4. Observed Test Statistic:
- Difference in proportions [tex]\( \hat{p}_1 - \hat{p}_2 \approx 0.657 - 0.615 = 0.042 \)[/tex]

5. Randomization Process:
- Combine both samples: total successes [tex]\( x_{\text{total}} = 359 + 359 = 718 \)[/tex]
- Total subjects [tex]\( n_{\text{total}} = 546 + 584 = 1130 \)[/tex]
- Under [tex]\( H_0 \)[/tex], success probability [tex]\( p_{\text{combined}} = \frac{718}{1130} \approx 0.635 \)[/tex]

6. Perform Simulations:
- Randomly assign 718 successes among 1130 trials 2000 times, given the seed value.
- For each assignment, calculate the difference in proportions.

7. Determine P-value:
- Count the proportion of simulated differences that are at least as extreme as the observed difference (0.042).

8. Decision Rule:
- If the [tex]\( P \)[/tex]-value is less than or equal to [tex]\( \alpha \)[/tex], reject [tex]\( H_0 \)[/tex].
- Otherwise, do not reject [tex]\( H_0 \)[/tex].

### Result:
Given a significance level of [tex]\( \alpha = 0.05 \)[/tex] and following the steps listed, the estimated P-value from the Randomization test for two proportions applet using 2000 random assignments with seed value 12641 is:

[tex]\[ P\text{-value} \approx 0.462 \][/tex]

Since [tex]\( 0.462 > 0.05 \)[/tex], we do not reject the null hypothesis [tex]\( H_0 \)[/tex]. This implies that there is not enough evidence to suggest that the proportion [tex]\( p_1 \)[/tex] is greater than the proportion [tex]\( p_2 \)[/tex].
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